The taxi takeoff time for commercial jets is a random region variable x with a m

Question

The taxi takeoff time for commercial jets is a random region variable x with a mean of 9 minutes and a standard deviation of 3.2 minutes. Assume that the distribution of taxi and take off times is approximately normal. You may assume that the jets are lined up on a runway so that one taxies and takes off immediately after the other, and that they take off one at a time on a given runway. (a) What is the probability that for 31 jets on a given runway, total taxi and takeoff time will be less than 320 minutes? (b) What is the probability that for 31 jets on a given runway, total taxi and takeoff time will be less than 275 minutes? (c) What is the probability that for 31 jets on a given runway, total taxi and takeoff time will be less than 275 and 320 minutes? In the following problem, check that it is appropriate to use the normal approximation to the binomial. Then Use the normal distribution to estimate the requested probabilities. More than a decade ago, high levels of lead in the blood put 80% of children at risk. A concerted effort was made to remove lead from the environment. Now suppose only 14% of children in the United States are at risk of high blood-lead levels. (a) In a random sample of 204 children taken more than a decade ago, what is the probability that 50 or more had high blood-lead levels? (b) In a random sample of 204 children taken now, what is the probability that 50 or more had high blood-lead levels? In the following problem, check that it is appropriate to use the normal approximation to the binomial. Then Use the normal distribution to estimate the requested probabilities. It is estimated that 3.5 % of the general population will live past their 90th birthday. In a graduating class of 746 high school seniors, find the following probabilities. (a) 15 or more will live beyond their 90th birthday (b)30 or more will live beyond their 90th birthday (c)between25 and 35 will live beyond their 90th birthday

Explanation / Answer

Normal Distribution
Mean ( u ) =9
Standard Deviation ( sd )=3.2
Normal Distribution = Z= X- u / sd ~ N(0,1)                  
a.
Taking an average 320 minutes = 320/31 = 10.3226

P(X < 10.3226) = (10.3226-9)/3.2
= 0.4133
= P ( Z <0.4133) From Standard Normal Table
= 0.660307   
b.
Taking an average 275 minutes = 275 /31 = 8.8710
P(X > 8.8710) = (8.8710-9)/3.2
= -0.0403
= P ( Z >-0.0403) From Standard Normal Table
= 0.516073

c.
P(7.432 < X < 8.649) = 0.660307-(1-0.516073 )=0.1764

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